Question

exit ticket

1. if you exert a force fₐ on an object, the force which the object exerts on

you will
a. depend on whether or not the object is moving.
b. depend on whether or not you are moving.
c. depend on the relative masses of you and the object.
d. always be the same magnitude as fₐ.

1. action - reaction forces...(select all that apply)

a. sometimes act on the same object.
b. always act on the same object.
c. are always equal in magnitude
d. are always of different magnitudes
e. can sometimes be of different magnitudes
f. may be at right angles.
g. always act on different objects.
h. always act in the same direction
i. always act in opposite directions

homework
a 12 kg box is sliding across a surface while experiencing a leftward applied
force of 100 n. the box is accelerating at - 1.0 m/s². draw a free - body diagram
showing all forces acting on the box and calculate the force of friction.

Explanation:

Question 1

This question is about Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. So when you exert a force \( F_a \) on an object, the object exerts a force on you that is equal in magnitude (and opposite in direction), regardless of motion or masses. Option a is wrong because motion doesn't affect the reaction force. Option b is wrong for the same reason. Option c is wrong as masses don't determine the magnitude of the reaction force (Newton's third law, not second). Option d is correct as per Newton's third law.

• Option a: Action - reaction forces act on different objects, so this is wrong.
• Option b: They act on different objects, so this is wrong.
• Option c: By Newton's third law, action - reaction forces are always equal in magnitude, so this is correct.
• Option d: They are equal in magnitude, so this is wrong.
• Option e: They are always equal, so this is wrong.
• Option f: Action - reaction forces can be at right angles (e.g., when a charged particle moves in a magnetic field, the magnetic force on the particle and the force the particle exerts on the magnetic field source can be at right angles), so this is correct.
• Option g: Action - reaction forces always act on different objects (action on one object, reaction on the other), so this is correct.
• Option h: They act in opposite directions, so this is wrong.
• Option i: By Newton's third law, they act in opposite directions, so this is correct.

Step 1: Recall Newton's second law

Newton's second law is \( \sum F = ma \), where \( \sum F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. The forces acting on the box are the applied force (\( F_{applied}=- 100\ N \), negative because it's to the left), the force of friction (\( F_f \), let's assume the positive direction is to the right, so friction will be to the right if the box is moving left? Wait, no, if the box is sliding across the surface and the applied force is leftward, and it's accelerating leftward (\( a=- 1.0\ m/s^2 \)), the friction force opposes the motion. Wait, the acceleration is \( - 1.0\ m/s^2 \) (leftward), mass \( m = 12\ kg \). The net force \( \sum F=F_{applied}+F_f \) (since applied force is left (\( - 100\ N \)) and friction: if the box is moving, friction is kinetic friction, opposing the motion. Wait, let's define the coordinate system: let right be positive. Then \( F_{applied}=- 100\ N \) (left), \( a=-1.0\ m/s^2 \) (left), \( m = 12\ kg \).

Step 2: Apply Newton's second law

\( \sum F=ma \)
\( F_{applied}+F_f=ma \)
We need to solve for \( F_f \).

Step 3: Substitute the values

\( - 100+F_f=12\times(- 1.0) \)
\( - 100+F_f=- 12 \)

Step 4: Solve for \( F_f \)

Add 100 to both sides: \( F_f=- 12 + 100=88\ N \)
(The positive sign indicates that the friction force is to the right, which makes sense as it opposes the leftward motion? Wait, no, if the box is accelerating leftward, the net force is leftward. The applied force is leftward (\( - 100\ N \)) and friction is rightward (\( + F_f \)). So \( - 100+F_f=12\times(- 1) \)
\( F_f=100 - 12 = 88\ N \). Wait, but if the net force is leftward (\( ma = 12\times(-1)=-12\ N \)), then \( F_{applied}+F_f=-12 \), \( - 100+F_f=-12 \), so \( F_f = 88\ N \). The free - body diagram will have:

• Gravitational force (\( F_g = mg=12\times9.8 = 117.6\ N \)) downward.
• Normal force (\( F_N \)) upward, equal in magnitude to \( F_g \) (since there's no acceleration in the vertical direction, \( \sum F_y = 0\), so \( F_N=F_g \)).
• Applied force (\( F_{applied} = 100\ N \)) to the left.
• Force of friction (\( F_f = 88\ N \)) to the right (opposing the motion, since the box is moving left? Wait, no, if the box is moving left and accelerating left, the net force is left. So the applied force (left) is greater than the friction force (right). So \( F_{applied}-F_f=ma \) (if we take left as positive). Let's re - define left as positive. Then \( F_{applied}=100\ N \), \( a = 1.0\ m/s^2 \), \( m = 12\ kg \). Then \( \sum F=F_{applied}-F_f=ma \) (because friction is opposite to motion, so rightward, negative in left - positive coordinate). So \( 100 - F_f=12\times1 \)

\( F_f=100 - 12 = 88\ N \). Yes, that's correct. So the friction force is 88 N to the right (if left is positive, it's - 88 N, but magnitude is 88 N).

Answer:

d. always be the same magnitude as \( F_a \)

Question 2